Part One: Solubility Equilibria. Insoluble and slightly soluble compounds are important in nature and commercially.

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CHAPTER 17: SOLUBILITY AND COMPLEX ION EQUILIBRIA Part One: Solubility Equilibria A. Ksp, the Solubility Product Constant. (Section 17.1) 1. Review the solubility rules. (Table 4.1) 2.

1 CHAPTER 17: SOLUBILITY AND COMPLEX ION EQUILIBRIA Part One: Solubility Equilibria A. Ksp, the Solubility Product Constant. (Section 17.1) 1. Review the solubility rules. (Table 4.1) 2. Insoluble and slightly soluble compounds are important in nature and commercially. a. b. c. 3. Chapter 17 Solubility behavior expressed mathematically by special equilib. constant = Ksp. a. BaSO4(s) Ba2+(aq) + SO42-(aq) b. Ca3(PO4)2 3 Ca PO43- Page 1

2 4. Note - in all saturated aqueous solutions of BaSO 4, no matter what other materials are present: 5. Table of K sp in Table Solubility = moles of a compound which dissolve in a liter of solution. Directly related to but not equal to K sp. B. Finding K sp. (Section 17.1) 1. Usually determined by conductivity experiment. 2. Example: Found that g of CaCO 3 can dissolve in 1.00 L of water. What is K sp of CaCO 3? 3. Example: The molar solubility of PbBr 2 is moles/liter. What is K sp? PbBr 2 (s) Pb 2+ (aq) + 2 Br - (aq) Chapter 17 Page 2

3 4. For Bi 2 S 3 with molar solubility = s C. Finding the Solubility from K sp. (Section 17.1) 1. Example: Calculate the molar solubility of AgCl and the concentrations [Ag + ] and [Cl - ] in a saturated aqueous solution of AgCl. The K sp = Example: Calculate the molar solubility of Bi 2 S 3, and [Bi 3+ ] and [S 2- ] concentrations. The K sp = K sp = [Bi 3+ ] 2 [S 2- ] 3 Bi 2 S 3 (s) 2 Bi S 2- solubility s 2s 3s Chapter 17 Page 3

4 D. The Common Ion Effect in Solubility Calculations. (Section 17.2) 1. Previously showed that in saturated solution of AgCl: [Cl - ] = [Ag + ] = M = K sp 2. Calculate the solubility of AgCl and the concentration of ionic species present in a 2.0 M NaCl solution saturated with AgCl. K sp of AgCl is Here Cl - ion is present from two sources: NaCl and AgCl. Chapter 17 Page 4

5 Part Two: Uses of the Solubility Product Principle A. Predicting if Precipitation Will Occur. (Section 17.3) 1. Precipitation will occur whenever the ion concentration product for a compound exceeds the K sp of that compound. For example, when [Ag + ][Cl - ] > K sp 2. The solution is unstable, and precipitate of AgCl will occur until: [Ag + ][Cl - ] = K sp 3. Example: The following reagents are added to water and the final volume is adjusted to 1.00 L. What, if any, compounds will precipitate? 1.0 g NaOH 40.0 g/mol = M 1.0 g Na 2 CO g/mol = M 1.0 g CaCl g/mol = M 1.0 g PbNO g/mol = M (Hint: Don t expect any Na + or NO 3 - salts to precipitate due to their high solubility.) Ca(OH) 2? [Ca 2+ ][OH - ] 2 = ( )( ) 2 = < K sp = NO! CaCO 3? [Ca 2+ - ][CO 3 ] = ( )( ) = > K sp = YES! Pb(OH) 2? [Pb 2+ ][OH - ] 2 = ( )( ) 2 = > K sp = YES! PbCO 3? [Pb 2+ - ][CO 3 ] = ( )( ) = > K sp = YES! PbCl 2? [Pb 2+ ][Cl - ] 2 = ( )( ) 2 = < K sp = NO! Pb(OH) 2 will precipitate first, because its ion product is the greatest relative to its K sp. Chapter 17 Page 5

6 B. Predicting Concentration Required to Cause Precipitation. (Section 17.3) 1. Example: A solution is M Na 2 CO 3. Then, calcium ions are added by another source. At what concentration of Ca 2+ ion will precipitation of CaCO 3 occur? K sp of CaCO 3 is Part 2: How many grams of CaCl 2 would need to be added to 500 ml of M Na 2 CO 3 to start forming this precipitate? precipitate when [Ca 2+ ] = M (moles/liter) OR: = moles in 500 ml moles g/mol = grams CaCl 2 C. Fractional Precipitation (Section 17.3) 1. May wish to precipitate some ions from solution but leave others. 2. Example: Have solution: 0.10 M Cu M Ag M Au + Slowly add NaCl to the solution. Which will precipitate first: CuCl, AgCl, or AuCl? At what [Cl - ] will these each begin to precipitate? Given: K sp of CuCl = K sp of AgCl = K sp of AuCl = Chapter 17 Page 6

7 3. What will [Au + ] be when AgCl just begins to precipitate? [Au + ][Cl - ] = M to precipitate AgCl [Au + ] = [Au + ] = M (remember, [Au + ] was originally 0.10 M ) D. Effect of ph on Solubility (Section 17.4) 1. Example: Limestone/Marble tends to dissolve in acidic conditions. 2. Solubility equilibrium for CaCO 3 (s) CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq) 3. This equilibrium is affected by addition of strong acid: H 3 O + (aq) + CO 3 2- (aq) H 2 O + HCO 3 - (aq) 4. This reaction removes carbonate ion, shifting the solubility equilibrium to the right, causing more CaCO 3 (s) to go into solution. A. Complex-Ion Formation. (Section 17.5) Part Four: Complex-Ion Equilibria 1. Some metal ions in solution can bind varying numbers of molecules or molecular ions. 2. Example: Ag + ion and ammonia molecules: Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) Ag(NH 3 ) 2 + (aq) Chapter 17 Page 7

3. The combined complexation reaction is then: Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) 4. This equilibrium is governed by the formation constant, or stability constant, K f.

8 3. The combined complexation reaction is then: Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) 4. This equilibrium is governed by the formation constant, or stability constant, K f. K f = [Ag(NH 3) + 2 ] = 1.7 [Ag ][NH 3 ] 5. The dissociation constant K d is simply the reciprocal of K f. 6. Calculations with this equilibrium. (See text Section 17.5) 7. Formation of complex ions can have important effects on solubility of ions in solution. (See lab manual - Experiments #21-26) Chapter 17 Page 8

Solubility Equilibria

Chapter 17 SOLUBILITY EQUILIBRIA (Part II) Dr. Al Saadi 1 Solubility Equilibria The concept of chemical equilibrium helps to predict how much of a specific ionic compound (salt) will dissolve in water.